Re: perl6 operator precedence table

2002-10-24 Thread Adam D. Lopresto
Really what I've been wishing for was an operator (or whatever) to let me do an s// without changing the variable. print 'He said '_($statement ~ s/\.$//)_', but we didn't believe him.'; I'm not sure exactly what the semantics would be, but somehow =~ without the = seems appealing...it's always

RE: perl6 operator precedence table

2002-10-24 Thread fearcadi
Damian Conway wrote: I certainly wouldn't mind seeing it return to that role, now that it's not needed elsewhere. And, of course, that would actually be: $x ~ $y string concatentation $x ~= $ystring append ~$x stringification ... $str =~

Literate programming (was Re: perl6 operator precedence table)

2002-10-24 Thread Trey Harris
Larry, As long as you're trying to figure out how to shoehorn in the last few available punctuation symbols, and thinking about if there are any bracketers left, I wondered if there was a chance of a chunking operator for literate programming? So you can do something like this, if were the

Re: perl6 operator precedence table

2002-10-24 Thread Michael Lazzaro
On Thursday, October 24, 2002, at 11:22 AM, Larry Wall wrote: But we also have to balance it against the desirability of using ~ for concatenation. Requiring whitespace around _ is a bit of a rationalization after the fact, and ~ escapes that problem in most cases. So (w/out whitespaces):

Re: perl6 operator precedence table

2002-10-24 Thread David Wheeler
On Thursday, October 24, 2002, at 10:34 AM, Larry Wall wrote: On the other hand, the current rule for recognizing the *end* of a name in the style of operator:=+ is to go till the next whitespace, on the assumption that we'll never have (shudder) whitespace operators. Oooh, I nominate

Re: perl6 operator precedence table

2002-10-24 Thread Luke Palmer
From: Angel Faus [EMAIL PROTECTED] Date: Fri, 25 Oct 2002 00:54:09 +0200 All this ones fit more with the concept of mystical analogy hinted by =~ than with the plain similarity that one would expect from like True. Can't say I like, um, like. Oh, and =~ looks much more intimidating,

Re: perl6 operator precedence table

2002-10-24 Thread Chris Dutton
Or we could go with Valspeak: $a is like $b and stuff At the moment I like like the best, actually... Hmmm... I could actually see like in a more active role. Along the lines of: my str $string; my $other_string is like $string; Analogous to saying: my str $other_string Except that

Re: perl6 operator precedence table

2002-10-24 Thread Larry Wall
On Thu, 24 Oct 2002, Chris Dutton wrote: : Also, this brings to mind the one thing I actually remember about : Sather, and as long as we're discussing operators... : : Will we have similar to Sather's ::=? That was essentially the : statically type this variable at run-time based on the type

Re: perl6 operator precedence table

2002-10-24 Thread Angel Faus
At the moment I like like the best, actually... like is beautiful for old-style regex matching, but I find it confusing for the new smart abilities: $varlike Class:Foo # $var is instance of Class:Foo $item like %hash # %hash{$item} is true $digit like (0..10) #

Re: perl6 operator precedence table

2002-10-24 Thread Larry Wall
On Thu, 24 Oct 2002, Damian Conway wrote: : Adam D. Lopresto wrote: : : Really what I've been wishing for was an operator (or whatever) to let me do an : s// without changing the variable. : : I would hope/expect that that's what the subroutine form of Cs would do. The problem with defining

RE: perl6 operator precedence table

2002-10-24 Thread Larry Wall
On Thu, 24 Oct 2002, fearcadi wrote: : Maybe , my question really is , how perl will behave if I will do : : sub operator:=+ (str $x, str $y) { system( $x | $y ) } ; : : so this is more question of qrammar ? The general rule in most lexers has always been that it grabs the longest token it can

Re: perl6 operator precedence table

2002-10-24 Thread Larry Wall
On Thu, 24 Oct 2002, Deborah Ariel Pickett wrote: : Which looks better? : if ($a == 1|2|3 || $b eq x|y|z) : or : if ($a == 1||2||3 | $b eq x||y||z) : ? I think disjunctions of data values should be | and disjunctions of expressions should be ||, so that the bigger concept has the bigger

Re: [OT] Power of Lisp macros?

2002-10-24 Thread Adriano Nagelschmidt Rodrigues
Luke Palmer writes: Do you think that Lisp macros make the language more powerful than others (eg Perl)? I mean, do they really give a competitive advantage, or are they being overrated (see below)? If you define powerful as can do more things, then of course not. No, of course. I guess

Re: perl6 operator precedence table

2002-10-24 Thread Michael Lazzaro
Brent Dax wrote: Can the new nefarious use be concat? Pretty please? On Wednesday, October 23, 2002, at 07:46 PM, Damian Conway wrote: I guess the only concern is the potential for nasty surprises between: $str =~ s/a/b/; substitute a for b in $str and: $str ~= s/a/b/; substitute

Re: perl6 operator precedence table

2002-10-24 Thread Smylers
Larry Wall wrote: On 20 Oct 2002, Smylers wrote: : Seems like not too long ago we were short of punctuation symbols, : and now you've got a spare one lying around. Pity there's no extra brackets lying around without going to Unicode... Well if C~ were made the hyper prefix (squiggly

Re: perl6 operator precedence table

2002-10-24 Thread Larry Wall
On Thu, 24 Oct 2002, Michael Lazzaro wrote: :$str1 ~ $str2# $str1 =~ m/$str2/ That would be a smart match, not m/$str2/. :$str ~ /foo/ # $str1 =~ m/foo/ That would work. :$str2 = ($str ~ /foo/bar/); # perform subst, assign result to $str2 : :

Re: perl6 operator precedence table

2002-10-24 Thread Jonathan Scott Duff
On Thu, Oct 24, 2002 at 09:59:00AM -0700, Michael Lazzaro wrote: Noone ever guesses that =~ means matching That's because it doesn't. =~ means something more akin to apply but it's only valid for the three m//, s///, tr/// ops. That'll change in perl 6 though :-) If anything, I'd almost

Re: perl6 operator precedence table

2002-10-24 Thread Austin Hastings
In 'C', we have: a = b+c; In Perl, we can have: $a = $b$c; (Parseable as $a = $b operator:spacespace operator:tab operator:spacespace $c;) Oh frabjous day! =Austin --- David Wheeler [EMAIL PROTECTED] wrote: On Thursday, October 24, 2002, at 10:34 AM, Larry Wall wrote: On

Re: perl6 operator precedence table

2002-10-24 Thread David Wheeler
On Thursday, October 24, 2002, at 02:52 PM, Austin Hastings wrote: In 'C', we have: a = b+c; In Perl, we can have: $a = $b$c; (Parseable as $a = $b operator:spacespace operator:tab operator:spacespace $c;) Oh frabjous day! Good Lord, you're sicker than I am! :-D David --

RE: perl6 operator precedence table

2002-10-24 Thread Shapiro, Jonathan
Um, I don't know about your mail program, but mine converts operator:tab to operator:spacespace operator:spacespace Anyone for makefiles? -Original Message- From: David Wheeler [mailto:david;wheeler.net] Sent: Thursday, October 24, 2002 5:59 PM To: [EMAIL

Re: [OT] Power of Lisp macros?

2002-10-24 Thread Adam Turoff
On Thu, Oct 24, 2002 at 12:26:41PM -0300, Adriano Nagelschmidt Rodrigues wrote: Luke Palmer writes: Lisp is implemented in C, and C's macros are certainly not essential to its functionality. But think of what macros in general provide: * Multi-platform compatability *

Re: perl6 operator precedence table

2002-10-24 Thread Larry Wall
On Fri, 25 Oct 2002, Martin D Kealey wrote: : Going back to Perl5 for a moment, we have : : substr($str,$start,$len) = $newstr : : why not simply extend pattern-matching in a similar way to substr, making it : an L-value, so that one gets : : $str ~ /[aeiou]+/ = vowels($) : : or : :